Question

Find the expansion of$\mathbf{(}\mathit{x}\mathbf{+}\mathit{y}{\mathbf{)}}^{\mathbf{4}}$$\mathbf{(}\mathit{x}\mathbf{+}\mathit{y}{\mathbf{)}}^{\mathbf{4}}$

a) using combinatorial reasoning, as in Example

b) using the binomial theorem.

Short answer

$x^4+4x^{3}y+6x^2{y}^2+4x{y}^3+y^4$

Step-by-step solution

Formula for Binomial theorem

Binomial theorem is:$\mathbf{(}\mathit{x}\mathbf{+}\mathit{y}{\mathbf{)}}^{\mathbf{n}}\mathbf{=}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathbf{ }\left(\begin{array}{l}n\\ j\end{array}\right){\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{j}}{\mathit{y}}^{\mathbf{j}}$

Use Distributive property and Binomial theorem

Given:$(x+y{)}^4$

(a)

Write the given statement as a product and then use the distributive property

$\begin{array}{r}(x+y{)}^4=(x+y\left)\right(x+y\left)\right(x+y\left)\right(x+y)\\ (x+y{)}^4=\left(x^2+xy+xy+y^2\right)\left(x^2+xy+xy+y^2\right)\end{array}$

$\begin{array}{r}(x+y{)}^4=\left(x^2+2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ (x+y{)}^4=x^4+2x^{3}y+x^2{y}^2+2x^{3}y+4x^2{y}^2+2x{y}^3+x^2{y}^2+2x{y}^3+y^4\\ (x+y{)}^4=x^4+4x^{3}y+6x^2{y}^2+4x{y}^3+y^4\end{array}$

(b)

Use the binomial theorem:

$(x+y{)}^4=\left(\begin{array}{l}4\\ 0\end{array}\right)x^4{y}^0+\left(\begin{array}{l}4\\ 1\end{array}\right)x^3{y}^1+\left(\begin{array}{l}4\\ 2\end{array}\right)x^2{y}^2+\left(\begin{array}{l}4\\ 3\end{array}\right)x^1{y}^3+\left(\begin{array}{l}4\\ 4\end{array}\right)x^0{y}^4$

$(x+y{)}^4=\frac{4!}{0!4!}{x}^4+\frac{4!}{1!3!}{x}^{3}y+\frac{4!}{2!2!}{x}^2{y}^2+\frac{4!}{3!1!}x{y}^3+\frac{4!}{4!0!}{y}^4$

$\begin{array}{r}(x+y{)}^4=1x^4+4x^{3}y+6x^2{y}^2+4x{y}^3+1y^4\\ (x+y{)}^4=x^4+4x^{3}y+6x^2{y}^2+4x{y}^3+y^4\end{array}$